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Chapter 4: Problem 40

On a multiple-choice exam with 3 possible answers for each of the 5 questions, what is the probability that a student will get 4 or more correct answers just by guessing?

Short Answer

Expert verified
The probability of a student getting 4 or more correct answers just by guessing on a 5-question multiple-choice exam with 3 possible answers for each question is \( \frac{11}{243} \).

Step by step solution

01

Identify the probability of guessing correctly and incorrectly

Since there are three possible answers for each question, and only one answer is correct, the probability of guessing an answer correctly is 1/3. Consequently, the probability of guessing an answer incorrectly is 2/3.
02

Calculate the probability of getting 4 correct answers

We will now use the binomial probability formula to find the probability of getting exactly 4 correct answers: P(X = 4) = C(5, 4) * (1/3)^4 * (2/3)^1 where C(5, 4) is the number of combinations of 5 questions taken 4 at a time: C(5, 4) = 5! / (4!(5-4)!) = 5 Now, substitute the values into the formula: P(X = 4) = 5 * (1/3)^4 * (2/3)^1 = 5 * (1/81) * (2/3) = 10/243
03

Calculate the probability of getting 5 correct answers

Next, we calculate the probability of guessing all answers correctly: P(X = 5) = C(5, 5) * (1/3)^5 * (2/3)^0 where C(5, 5) is the number of combinations of 5 questions taken 5 at a time: C(5, 5) = 5! / (5!(5-5)!) = 1 Now, substitute the values into the formula: P(X = 5) = 1 * (1/3)^5 * (2/3)^0 = 1 * (1/243) * 1 = 1/243
04

Calculate the probability of getting 4 or more correct answers

Now we will sum up the probabilities of getting 4 and 5 correct answers: P(X >= 4) = P(X = 4) + P(X = 5) = 10/243 + 1/243 = 11/243 So, the probability of a student getting 4 or more correct answers just by guessing is 11/243.

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Most popular questions from this chapter

One of the numbers 1 through 10 is randomly chosen. You are to try to guess the number chosen by asking questions with "yes-no" answers. Compute the expected number of questions you will need to ask in each of the following two cases: (a) Your ith question is to be "Is it i?" \(i=\) 1,2,3,4,5,6,7,8,9,10 (b) With each question you try to eliminate onehalf of the remaining numbers, as nearly as possible.

Four buses carrying 148 students from the same school arrive at a football stadium. The buses carry, respectively, \(40,33,25,\) and 50 students. One of the students is randomly selected. Let \(X\) denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let \(Y\) denote the number of students on her bus. (a) Which of \(E[X]\) or \(E[Y]\) do you think is larger? Why? (b) Compute \(E[X]\) and \(E[Y]\)

A total of \(2 n\) people, consisting of \(n\) married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let \(C_{i}\) denote the event that the members of couple \(i\) are seated next to each other, \(i=1, \ldots, n\) (a) Find \(P\left(C_{i}\right)\) (b) For \(j \neq i,\) find \(P\left(C_{j} | C_{i}\right)\) (c) Approximate the probability, for \(n\) large, that there are no married couples who are seated next to each other.

In the game of Two-Finger Morra, 2 players show 1 or 2 fingers and simultancously guess the number of fingers their opponent will show. If only one of the players guesses correctly, he wins an amount (in dollars) equal to the sum of the fingers shown by him and his opponent. If both players guess correctly or if neither guesses correctly, then no money is exchanged. Consider a specificd player, and denote by \(X\) the amount of money he wins in a single game of Two-Finger Morra. (a) If each player acts independently of the other, and if each player makes his choice of the number of fingers he will hold up and the number he will guess that his opponent will hold up in such a way that each of the 4 possibilities is equally likely, what are the possible values of \(X\) and what are their associated probabilities? (b) Suppose that each player acts independently of the other. If each player decides to hold up the same number of fingers that he guesses his opponent will hold up, and if each player is equally likely to hold up 1 or 2 fingers, what are the possible values of \(X\) and their associated probabilities?

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or an "off" day. He figures that if he has an on day, then each of his examiners will pass him, independently of each other, with probability \(.8,\) whereas if he has an off day, this probability will be reduced to \(.4 .\) Suppose that the student will pass the examination if a majority of the examiners pass him. If the student feels that he is twice as likely to have an off day as he is to have an on day, should he request an examination with 3 examiners or with 5 examiners?
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